If 2N force is applied on 2 kg mass due east and same magnitude of force due west, thechange in velocity of the body in 2 sec is

Write a rule for the sequence. 3, -3, -9, -15. A. Start with 3 and add -6 repeatedly B. Start with -6 and add 3 repeatedly C. St

faust18 [17]

Substract two consecutive terms of the sequence to see if there is a common difference:

$\begin{gathered} (-3)-(3)=-3-3=-6 \\ (-9)-(-3)=-9+3=-6 \\ (-15)-(-9)=-15+9=-6 \end{gathered}$

As we can see, there is a common difference of -6.

Then, if a number of the sequence is given, the next one can be found by adding -6 (which is the same as subtracting 6).

Notice that the first term of the sequence is 3.

Then, the rule for the sequence is to start with 3 and add -6 repeatedly.

Therefore, the correct choice is option A) Start with 3 and add -6 repeatedly.

7 0

2 years ago

The average period of pendulum clock is found to be 1.2s at sea level. The period of the same pendulum on a mountain top is foun

Kipish [7]

Answer:

g' = 10.12m/s^2

Explanation:

In order to calculate the acceleration due to gravity at the top of the mountain, you first calculate the length of the pendulum, by using the information about the period at the sea level.

You use the following formula:

$T=2\pi \sqrt{\frac{l}{g}}$ (1)

l: length of the pendulum = ?

g: acceleration due to gravity at sea level = 9.79m/s^2

T: period of the pendulum at sea level = 1.2s

You solve for l in the equation (1):

$l=\frac{gT^2}{4\pi^2}\\\\l=\frac{(9.79m/s^2)(1.2s)^2}{4\pi^2}=0.35m$

Next, you use the information about the length of the pendulum and the period at the top of the mountain, to calculate the acceleration due to gravity in such a place:

$T'=2\pi \sqrt{\frac{l}{g'}}\\\\g'=\frac{4\pi^2l}{T'^2}$

g': acceleration due to gravity at the top of the mountain

T': new period of the pendulum

$g'=\frac{4\pi^2(0.35m)}{(1.18s)^2}=10.12\frac{m}{s^2}$

The acceleration due to gravity at the top of the mountain is 10.12m/s^2

5 0

3 years ago

Particle 1 and particle 2 have masses of m1 = 2.2 × 10-8 kg and m2 = 4.8 × 10-8 kg, but they carry the same charge q. The two pa

Lorico [155]

Answer: $r_2=11.81 cm$

Explanation:

Given

$m_1=2.2\times 10^{-8} kg$

$m_2=4.8\times 10^{-8} kg$

same charge on both masses

potential Energy due to Magnetic Field =Kinetic Energy of Particle

$qV=\frac{mv^2}{2}$

$v=\sqrt{\frac{2qV}{m}}$

and we know

Force due to magnetic field will Provide centripetal Force

$qvB=\frac{mv^2}{r}$

$B=\frac{\sqrt{\frac{2Vm}{q}}}{r}$

and B is equal for both particles

thus $\frac{m}{r^2}=constant$

$\frac{m_1}{r_1^2}=\frac{m_2}{r_2^2}$

$r_2^2=\frac{4.8}{2.2}\times 8^2$

$r_2=11.81 cm$

4 0

3 years ago

Displacement vectors of 4km north, 2km south, 5km north, 5km south combine to a total displacement of

goldfiish [28.3K]

<u>Answer</u>

The combined displacement is 2km north

<u>Explanation</u>

Since displacement is a vector quantity, we take into account the direction.

Good for us all the displacement vectors are in the same dimension, so we can make north positive and south negative or vice-versa.

We now add to obtain,

$4+-2+5+-5$

This will simplify to

$=4-2+5-5=2$

Therefore the combined displacement is 2km north

5 0

3 years ago

A 4 kg textbook sits on a desk. It is pushed horizontally with a 50 N applied force against a 15 N frictional force.

GarryVolchara [31]

a) See free-body diagram in attachment

b) The book is stationary in the vertical direction

c) The net horizontal force is 35 N in the forward direction

d) The net force on the book is 35 N in the forward horizontal direction

e) The acceleration is $8.75 m/s^2$ in the forward direction

Explanation:

a)

The free-body diagram of a body represents all the forces acting on the body using arrows, where the length of each arrow is proportional to the magnitude of the force and points in the same direction.

From the diagram of this book, we see there are 4 forces acting on the book:

- The applied force, F = 50 N, pushing forward in the horizontal direction

- The frictional force, $F_f = 15 N$ , pulling backward in the horizontal direction (the frictional force always acts in the direction opposite to the motion)

- The weight of the book, $W=mg$ , where m is the mass of the book and $g=9.8 m/s^2$ is the acceleration of gravity, acting downward. We can calculate its magnitude using the mass of the book, m = 4 kg:

$W=(4)(9.8)=39.2 N$

- The normal reaction exerted by the desk on the book, N, acting upward, and balancing the weight of the book

b)

The book is in equilibrium in the vertical direction, therefore there is no motion.

In fact, the magnitude of the normal reaction (N) exerted by the desk on the book is exactly equal to the weight of the book (W), so the equation of motion along the vertical direction is

$N-W=ma$